y=t/t+1 t2-C2at+4c2=0 判别式>=0, c4a2-16c2>=0 c2a2>=16 (C≠0)、所以 IacI≥4 2.f(x+1)-f(x)=x+1/x+2-x/1+x=[(x+1)^2-x(x+2)]/(x+1)(x+2)=1/(x+1)(x+2)>0 ( x+1)>=0)fx在(-1,正无穷大)上f(x)单增